23,244 views
54 votes
54 votes
Consider a $128 \times 10^3$ bits/second satellite communication link with one way propagation delay of $150$ milliseconds. Selective retransmission (repeat) protocol is used on this link to send data with a frame size of $1$ kilobyte. Neglect the transmission time of acknowledgement. The minimum number of bits required for the sequence number field to achieve $100 \%$ utilization is ________.

6 Answers

Best answer
90 votes
90 votes

Answer is 4 bits.

As  we want $100 \%$ efficiency$(\mu),\ \large {w_s}\geq 1+2a$

 $a = \dfrac{\text{propagation time}}{\text{transmission time}}$

$\quad=\dfrac{150  }{1024 \times \dfrac{8}{128}} = \dfrac{150}{64} =  2.34,$

 $\Rightarrow \large {w_s} \geq 1+2a = \lceil  5.6875 \rceil =  6$

Available $\text{seq numbers} \geq \large w_s+w_r$

In Selective Repeat,

$\large w_s=w_r$ (let it be n)

$2\times n=2\times 6=12$

avail $\text{seq numbers}\geq 12$

So, minimum $\text{seq numbers}$ are $12$.

Number of bits for that is $\lceil \log_2 {12} \rceil =4.$

edited by
18 votes
18 votes
Transmission time of a frame = (10^3 * 8 * 10^3)/ (128 * 10^3) = 62.5 ms.

RTT = 150* 2 ms = 300 ms

(One frame transmission time + RTT) is the maximum time that sender can transmit without waitng for acknowledgement.

Here utilization is 100% i.e. efficiency is 1

So in (62.5 +300) ms sender can send atmost 362.5/ 62.5 frame = 5.8 i.e approx 6 frames.

In selective repet, sender window size = receiver window size. So window size = 6 * 2 = 12.

So, 4 bits to represent window.
7 votes
7 votes
for 100% utilization sender can keep sending data till it receives an +ve or -ve acknowldegement.

So it can send for 150*2 ms

so total data that can be transmitted = 300*128 = 38400/8 = 4800bytes

4800/1024 = 4.6875 =  5 frames = 3 bits

This is because receiver can demand retransmission of first frame itself.
7 votes
7 votes

For selective repeat: size of window( W )  = 2k-1

(Utilization or Efficiency)selective repeat = $W \div ( 1 + RTT / TD)$

Since utilization is given 100%,

TD = 8 * 103 / 128 * 10

      = 8 / 128 sec

     = 0.0625 sec

RTT = 2 * PD 

        = 2 * 150 ms = 300 * 10-3 = 0.3 sec

1 = W / (1 + 0.3 / 0.0625)

W = 5.8 or,

2k-1 = 5.8

k = 4 only this value satisfies.

Thus the min no. of sequence bits required for 100% utilization is 4.

Answer:

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